At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
@Mani_Jha , @saifoo.khan , @heena , @supercrazy92 please help.
No chem! :(
Try at least.
@Mani_Jha is a boss at chem.
Yeah, that's why i called Mani Bhai also. He is my teacher. I hope he will do it this time too.:)))))
Let the concentration of NH3 be x. Now, the resultant solution was basic(as it had to be neutralized by NaOH). During the reaction of neutralization: M1V1=M2V2 x20=30*0.1 Find z. Let the initial volume of NH3 be V. Then when NH3 reacted initially with H2S04, 100 ml of H2S04 neutralized (v-20) ml of NH3(Because 20 ml was un-neutralized and made the solution basic) Again use the law of equivalents in this case(be careful: the n-factor of H2S04 is 2) Did you understand? @Saifoo.khan, you just guessed that, didnt u?
Mani Bhai, can u show the second equation what we need to take?
Okay, mani bhaiiiiiii
Wordings of these problems are terrible; teachers should bear in mind that students are not reading their minds and that wordings should be unambiguous. In first problem, I find 17 millimole NH3 or 0.38 litre in STP. That holds if "equivalence" in the wording is meant for 'first equivalence', because eventually, NH4+ formed will also be titrated. In second problem there should be 1.5 millimole of carbonate and 0.5 millimole of hydrogencarbonate. Here again, problem should read: "The second end point was reached when AN EXTRA 20 ml of the acid was added
Oooookkkkkk, I will consult with my teacher about this. But these r the exact wordings that our teacher gave.
What about the answers?
btw, do you use 22.4 litres/mole or 22.7 litres/mole for molar volume in STP?
Do you have expected answers ?
Not now. i need to submit it as my assignment. i couldn't do it, so i asked it here.
Do my answers coincide with yours of Mani-Jha's? Do you understand what reactions are at stake here?
My teacher gave the answer just now as 112 ml for the first one, and for the second one: Weight of NaOH = 0.8g, weight of Na2CO3 = 1.59 g
1st one is correct. I forgot to divide something by 5, as you take only a 20 mL sample of the 100 mL solution. So there were 5 millimoles of ammonia ie a volume of 112 mL
Thanks a lot
I have 159 mg of Na2CO3 which is not 1.59 g, unless concentration af acid is 1M instead of 0.1M. Then I do not understand the NaOH part. Your solution is supposed to have NaHCO3. What do you think?
Even I think the same.